Problem: Is ${14550}$ divisible by $9$ ?
Answer: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {14550}= &&{1}\cdot10000+ \\&&{4}\cdot1000+ \\&&{5}\cdot100+ \\&&{5}\cdot10+ \\&&{0}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {14550}= &&{1}(9999+1)+ \\&&{4}(999+1)+ \\&&{5}(99+1)+ \\&&{5}(9+1)+ \\&&{0} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {14550}= &&\gray{1\cdot9999}+ \\&&\gray{4\cdot999}+ \\&&\gray{5\cdot99}+ \\&&\gray{5\cdot9}+ \\&& {1}+{4}+{5}+{5}+{0} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first four terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${14550}$ is divisible by $9$ if ${ 1}+{4}+{5}+{5}+{0}$ is divisible by $9$ Add the digits of ${14550}$ $ {1}+{4}+{5}+{5}+{0} = {15} $ If ${15}$ is divisible by $9$ , then ${14550}$ must also be divisible by $9$ ${15}$ is not divisible by $9$, therefore ${14550}$ must not be divisible by $9$.